Integrand size = 27, antiderivative size = 121 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {5 d \sqrt {c+d x^3}}{96 c \left (8 c-d x^3\right )}-\frac {\sqrt {c+d x^3}}{24 x^3 \left (8 c-d x^3\right )}+\frac {3 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{128 c^{3/2}}-\frac {7 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{3/2}} \]
3/128*d*arctanh(1/3*(d*x^3+c)^(1/2)/c^(1/2))/c^(3/2)-7/384*d*arctanh((d*x^ 3+c)^(1/2)/c^(1/2))/c^(3/2)+5/96*d*(d*x^3+c)^(1/2)/c/(-d*x^3+8*c)-1/24*(d* x^3+c)^(1/2)/x^3/(-d*x^3+8*c)
Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {4 \sqrt {c} \left (4 c-5 d x^3\right ) \sqrt {c+d x^3}}{-8 c x^3+d x^6}+9 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )-7 d \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{384 c^{3/2}} \]
((4*Sqrt[c]*(4*c - 5*d*x^3)*Sqrt[c + d*x^3])/(-8*c*x^3 + d*x^6) + 9*d*ArcT anh[Sqrt[c + d*x^3]/(3*Sqrt[c])] - 7*d*ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/( 384*c^(3/2))
Time = 0.26 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.12, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {948, 109, 27, 168, 27, 174, 73, 219, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx\) |
| \(\Big \downarrow \) 948 |
| \(\displaystyle \frac {1}{3} \int \frac {\left (d x^3+c\right )^{3/2}}{x^6 \left (8 c-d x^3\right )^2}dx^3\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle \frac {1}{3} \left (-\frac {\int -\frac {c d \left (19 d x^3+28 c\right )}{2 x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3}{8 c}-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \int \frac {19 d x^3+28 c}{x^3 \left (8 c-d x^3\right )^2 \sqrt {d x^3+c}}dx^3-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 168 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}-\frac {\int -\frac {18 c d \left (5 d x^3+14 c\right )}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{72 c^2 d}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\int \frac {5 d x^3+14 c}{x^3 \left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 174 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {7}{4} \int \frac {1}{x^3 \sqrt {d x^3+c}}dx^3+\frac {27}{4} d \int \frac {1}{\left (8 c-d x^3\right ) \sqrt {d x^3+c}}dx^3}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 73 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {27}{2} \int \frac {1}{9 c-x^6}d\sqrt {d x^3+c}+\frac {7 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {7 \int \frac {1}{\frac {x^6}{d}-\frac {c}{d}}d\sqrt {d x^3+c}}{2 d}+\frac {9 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {1}{3} \left (\frac {1}{16} d \left (\frac {\frac {9 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}-\frac {7 \text {arctanh}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{2 \sqrt {c}}}{4 c}+\frac {5 \sqrt {c+d x^3}}{2 c \left (8 c-d x^3\right )}\right )-\frac {\sqrt {c+d x^3}}{8 x^3 \left (8 c-d x^3\right )}\right )\) |
(-1/8*Sqrt[c + d*x^3]/(x^3*(8*c - d*x^3)) + (d*((5*Sqrt[c + d*x^3])/(2*c*( 8*c - d*x^3)) + ((9*ArcTanh[Sqrt[c + d*x^3]/(3*Sqrt[c])])/(2*Sqrt[c]) - (7 *ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]])/(2*Sqrt[c]))/(4*c)))/16)/3
3.5.16.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/((m + 1)*(b*c - a*d)*(b*e - a*f))), x] + S imp[1/((m + 1)*(b*c - a*d)*(b*e - a*f)) Int[(a + b*x)^(m + 1)*(c + d*x)^n *(e + f*x)^p*Simp[(a*d*f*g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a* h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && ILtQ[m, -1]
Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))* ((c_.) + (d_.)*(x_))), x_] :> Simp[(b*g - a*h)/(b*c - a*d) Int[(e + f*x)^ p/(a + b*x), x], x] - Simp[(d*g - c*h)/(b*c - a*d) Int[(e + f*x)^p/(c + d *x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_. ), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^ p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ [b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Time = 4.64 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.82
| method | result | size |
| pseudoelliptic | \(\frac {d \left (-\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right ) d \,x^{3}+2 \sqrt {d \,x^{3}+c}\, \sqrt {c}}{2 d \,x^{3} c^{\frac {3}{2}}}+\frac {\frac {9 \sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {9 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{2 \sqrt {c}}}{c}\right )}{192}\) | \(99\) |
| risch | \(-\frac {\sqrt {d \,x^{3}+c}}{192 c \,x^{3}}-\frac {d \left (\frac {7 \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3 \sqrt {c}}-\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}-6 c \left (-\frac {\sqrt {d \,x^{3}+c}}{c \left (d \,x^{3}-8 c \right )}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{3 c^{\frac {3}{2}}}\right )\right )}{128 c}\) | \(114\) |
| default | \(\frac {-\frac {c \sqrt {d \,x^{3}+c}}{3 x^{3}}+\frac {2 d \sqrt {d \,x^{3}+c}}{3}-\sqrt {c}\, d \,\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{64 c^{2}}+\frac {d \left (\frac {2 d \,x^{3} \sqrt {d \,x^{3}+c}}{9}+\frac {8 c \sqrt {d \,x^{3}+c}}{9}-\frac {2 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{\sqrt {c}}\right )}{3}\right )}{256 c^{3}}+\frac {d \left (\frac {2 \sqrt {d \,x^{3}+c}}{3}-3 c \left (-\frac {\sqrt {d \,x^{3}+c}}{-d \,x^{3}+8 c}+\frac {\operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )}{\sqrt {c}}\right )\right )}{64 c^{2}}+\frac {d \left (81 c^{\frac {3}{2}} \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{3}+c}}{3 \sqrt {c}}\right )-\left (d \,x^{3}+28 c \right ) \sqrt {d \,x^{3}+c}\right )}{1152 c^{3}}\) | \(219\) |
| elliptic | \(\text {Expression too large to display}\) | \(1550\) |
1/192*d*(-1/2*(7*arctanh((d*x^3+c)^(1/2)/c^(1/2))*d*x^3+2*(d*x^3+c)^(1/2)* c^(1/2))/d/x^3/c^(3/2)+9*((d*x^3+c)^(1/2)/(-d*x^3+8*c)+1/2*arctanh(1/3*(d* x^3+c)^(1/2)/c^(1/2))/c^(1/2))/c)
Time = 0.41 (sec) , antiderivative size = 280, normalized size of antiderivative = 2.31 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\left [\frac {9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} + 6 \, \sqrt {d x^{3} + c} \sqrt {c} + 10 \, c}{d x^{3} - 8 \, c}\right ) + 7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {c} \log \left (\frac {d x^{3} - 2 \, \sqrt {d x^{3} + c} \sqrt {c} + 2 \, c}{x^{3}}\right ) - 8 \, {\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{768 \, {\left (c^{2} d x^{6} - 8 \, c^{3} x^{3}\right )}}, \frac {7 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{c}\right ) - 9 \, {\left (d^{2} x^{6} - 8 \, c d x^{3}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {d x^{3} + c} \sqrt {-c}}{3 \, c}\right ) - 4 \, {\left (5 \, c d x^{3} - 4 \, c^{2}\right )} \sqrt {d x^{3} + c}}{384 \, {\left (c^{2} d x^{6} - 8 \, c^{3} x^{3}\right )}}\right ] \]
[1/768*(9*(d^2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 + 6*sqrt(d*x^3 + c)*sqr t(c) + 10*c)/(d*x^3 - 8*c)) + 7*(d^2*x^6 - 8*c*d*x^3)*sqrt(c)*log((d*x^3 - 2*sqrt(d*x^3 + c)*sqrt(c) + 2*c)/x^3) - 8*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/(c^2*d*x^6 - 8*c^3*x^3), 1/384*(7*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arc tan(sqrt(d*x^3 + c)*sqrt(-c)/c) - 9*(d^2*x^6 - 8*c*d*x^3)*sqrt(-c)*arctan( 1/3*sqrt(d*x^3 + c)*sqrt(-c)/c) - 4*(5*c*d*x^3 - 4*c^2)*sqrt(d*x^3 + c))/( c^2*d*x^6 - 8*c^3*x^3)]
Timed out. \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\text {Timed out} \]
\[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\int { \frac {{\left (d x^{3} + c\right )}^{\frac {3}{2}}}{{\left (d x^{3} - 8 \, c\right )}^{2} x^{4}} \,d x } \]
Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.94 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {7 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{\sqrt {-c}}\right )}{384 \, \sqrt {-c} c} - \frac {3 \, d \arctan \left (\frac {\sqrt {d x^{3} + c}}{3 \, \sqrt {-c}}\right )}{128 \, \sqrt {-c} c} - \frac {5 \, {\left (d x^{3} + c\right )}^{\frac {3}{2}} d - 9 \, \sqrt {d x^{3} + c} c d}{96 \, {\left ({\left (d x^{3} + c\right )}^{2} - 10 \, {\left (d x^{3} + c\right )} c + 9 \, c^{2}\right )} c} \]
7/384*d*arctan(sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 3/128*d*arctan(1/3 *sqrt(d*x^3 + c)/sqrt(-c))/(sqrt(-c)*c) - 1/96*(5*(d*x^3 + c)^(3/2)*d - 9* sqrt(d*x^3 + c)*c*d)/(((d*x^3 + c)^2 - 10*(d*x^3 + c)*c + 9*c^2)*c)
Time = 8.21 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+d x^3\right )^{3/2}}{x^4 \left (8 c-d x^3\right )^2} \, dx=\frac {\frac {9\,d\,\sqrt {d\,x^3+c}}{32}-\frac {5\,d\,{\left (d\,x^3+c\right )}^{3/2}}{32\,c}}{3\,{\left (d\,x^3+c\right )}^2-30\,c\,\left (d\,x^3+c\right )+27\,c^2}+\frac {d\,\left (\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{\sqrt {c^3}}\right )\,1{}\mathrm {i}-\frac {\mathrm {atanh}\left (\frac {c\,\sqrt {d\,x^3+c}}{3\,\sqrt {c^3}}\right )\,9{}\mathrm {i}}{7}\right )\,7{}\mathrm {i}}{384\,\sqrt {c^3}} \]